Question

The instantaneous rate of disappearance of $MnO_4$ ion in the following reaction is $4.56 \times 10^{-3} \,Ms^{-1}$
$2MnO_{4}^{-} + 10I^{-}+ 16H^{+} \to 2Mn^{2+} + 5I_{2} + 8H_{2}O$
The rate of appearance $I_2$ is :

• A. $4.56 \times 10^{-4}\, Ms^{-1}$
• B. $1.14 \times 10^{-2}\, Ms^{-1}$
• C. $1.14 \times 10^{-3}\, Ms^{-1}$
• D. $5.7 \times 10^{-3}\, Ms^{-1}$

Answer 1

Answer: (B)

Given $- \frac{dMnO^{-}_{4}}{dt} = 4.56 \times 10^{-3}\,Ms^{-1}$
From the reaction given,
$-\frac{1}{2} \frac{dMnO^{-}_{4}}{dt} =\frac{ 4.56 \times 10^{-3}}{2}Ms^{-1}$
$-\frac{1}{2}\frac{dMnO^{-}_{4}}{dt} = \frac{1}{5} \frac{dI_{2}}{dt}$
$\therefore \quad- \frac{5}{2} \frac{dMnO^{-}_{4}}{dt} = \frac{dI_{2}}{dt}$
On substituting the given value
$\therefore \quad\frac{dI_{2}}{dt} =\frac{ 4.56 \times 10^{-3}\times5}{2} = 1.14\times10^{-2}M/s$